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As it can be seen in below schematics, I have a transformer connected to mains 220V and converts to 9V, which then connects to a bridge rectifier and voltage regulator through AC1 and AC2 labels.

This is a TRIAC control project and I am using the signal labeled SYNC to let the microcontroller know about the zero-crossing points.

I am carrying AC1 and AC2 to the bridge rectifer from an another board through some breakaway connectors and a PCB, which in total is about 8 cm long trace per signal.

Do I need to put a small valued (about 100nF) capacitor across AC1 and AC2? If so,where should I place this capacitor on the PCB, closer to the bridge rectifer or the transformer? How to calculate the voltage rating of the capacitor? What is the value of the capacitor needed?

What would be the purpose of this capacitor, in detail?

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I have not given a lot of details about R1 and C1 in the second schematic. I will try to give more information with this edit.

I am using D1 to isolate the filtered (DC) voltage so that I can have a zero-crossing detection signal. I wanted to have a 100 Hz, where my line frequency is 50 Hz. That's why I've taken the signal after the bridge rectifier.

Here are some scope-screen-shots of various configurations.

enter image description here enter image description here enter image description here

Secondary of the transformer*:

  1. R1 is populated, 7805 sourcing ~10mA
  2. R1 is populated, 7805 sourcing ~75mA
  3. R1 is not populated, 7805 sourcing ~75mA

enter image description here enter image description here enter image description here

Output of the bridge rectifier*:

  1. R1 is populated, 7805 sourcing ~10mA
  2. R1 is populated, 7805 sourcing ~75mA
  3. R1 is not populated, 7805 sourcing ~75mA

enter image description here enter image description here enter image description here

Input of 7805*:

  1. R1 is populated, 7805 sourcing ~10mA
  2. R1 is populated, 7805 sourcing ~75mA
  3. R1 is not populated, 7805 sourcing ~75mA

enter image description here

Signal labeled SYNC_In - R1 is populated, 7805 sourcing ~10mA

As you can see, in the scope-shot captioned Output of the bridge rectifier - R1 is not populated, 7805 sourcing ~75mA, voltage never goes down enough to create a SYNC_In signal.

*:Image captions are from left to right and are listed below the related images. Clicking on the image caption will open up bigger resolution image in the same tab.

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People don't normally put a cap on the secondary side of a regular ac transformer. What gave you the feeling one is required? –  Andy aka Dec 13 '13 at 13:22
@Andyaka My boss (dad) told me to :) –  abdullah kahraman Dec 13 '13 at 13:42
Tell him I said that both R1 and C1 in the lower schematic (low voltage section) are silly and that he is engaging in superstition-based design. –  Olin Lathrop Dec 13 '13 at 14:29
@MattYoung: Does too!! Without D1 the SYNC signal will be a smoothed DC because of C2. With D1 present SYNC will be the rectified sine from rectifier D2. –  The Resistance Dec 13 '13 at 18:45
Usually the fuse is placed inline with the Phase/Hot terminal. That way, if there is a fault downstream, the live power is cut. In your configuration, if there is a fault, the circuit still remains at live potential even if the fuse blows. –  Techie Dec 31 '13 at 20:45

3 Answers 3

R1 and C1 in the low voltage section (it would be nice if your component designators were unique accross all the schematic sections you post) are pointless.

R1 puts unnecessary load on the transformer output and has no useful purpose.

C1 will filter some high frequencies that may come in from the high voltage feed and make it accross the transformer, but there is no point to this. Only the positive peaks of the output of the secondary will conduct thru D1, which causes C2 to be charged up. A little high freqency noise superimposed on the power waveform would only make the peaks slightly higher, which is harmless. Furthermore, the inductance of the transformer will filter out much of the high frequency content anyway.

C1 in the high voltage circuit does make sense since it will reduce some of the high frequencies produced by this circuit and thereby fed back onto the power line. The half wave bridge rectifier will draw current in short spurts, which will contain significant harmonics. The inductance of the transformer together with C1 will reduce the magnitude of the higher harmonics that will be coupled back onto the power line.

For such a low power (tranformer is rated for only 2 VA), a simple parallel capacitor like C1 may be sufficient. For higher power levels a more deliberate filter may be required. For even higher power levels, a certain minimum power factor must be met, depending on what jurisdiction the device will be used or sold in.

However, all this is noise compared to another glaring problem with this circuit. The 9 V RMS output of the transformer will be half-wave rectified to produce about 12 V. So far so good. From the 2 VA rating of the transformer, we know that this supply is intended to supply less than 2 W. 2 W at 12 V would be 167 mA, so let's be generous and say it is intended to provide 100 mA only. Note that C1 will only be charged up once per power line cycles. Judging from the 230 V input and your location, we can assume this is at 50 Hz. That means C2 will be charged up every 20 ms.

For simplicity for now, let's see how much the voltage on C2 will drop assuming it gets charged to 12 V instantaneously once every 20 ms. (100 mA)(20 ms) / 220 µF = 9 V, which means the "12 V" line would drop to 3 V, which is clearly no good for the input of a 7805 regulator.

We can work this backwards and see what current this supply can maintain. Let's say the 7805 needs at least 7.5 V in to work properly, which mean C2 can drop no more than 4.5 V. (4.5 V)(220 µF) / 20 ms = 20 mA. Of course C2 doesn't get charged up instantaneously, so the actual value is a little higher. But still, this supply as shown is basically no good for output currents beyond 20 mA.


I just noticed that there is a bridge rectifier immediately on the output of the transformer. Somehow my brain skipped over that originally since I saw D1 there, which appeared to be a half wave rectifier.

Having both the bridge rectifier and D1 is more silliness. Because of the small value of C2, I'd keep the bridge and get rid of D1. This means the output will be charged up twice per power line cycle, but with one more diode drop in series. Let's say the total voltage drop of the bridge is 1.4 V. If the transformer can be counted on to produce 9 V sine, then its peaks will be 12.7 V. That minus the diode drops is 11.3 V. That leaves 3.8 V headroom for the 7805 to work. (3.8 V)(220 µF) / 10 ms = 84 mA, which is the maximum this supply can be counted on to produce from the C2 voltage drop point of view. (84 mA)(9 V) = 760 mW, so that should be within the capabilities of the 2 VA transformer.

If you need more than 80 mA or so from this supply, then you have to change something. This is assuming already that D1 is replaced by a short.

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D1 is necessary to ensure that there in an AC component at the SYNC terminal. Without D1, you would be trying to get the SYNC signal from the raw 12 volt supply. (Although there will be some ripple there, which might be enough to provide the SYNC signal.) –  Peter Bennett Dec 13 '13 at 17:26
@Peter: That's a silly way to produce a sync signal. Tie the voltage divider that feeds the base of the transistor to the AC1 or AC2 line. Each of those are guaranteed to go from a little below ground to a little above the "12 V" supply each line cycle. –  Olin Lathrop Dec 13 '13 at 17:30
Sorry for the designators. This is a multi-board project and the best way to achieve this in my CAD program (KiCad) is by creating separate project for each board. But I should have edited this out, then post.. Also, I need to sense the zero crossings at 100Hz, that's why I've put D1. –  abdullah kahraman Dec 14 '13 at 11:18
Thank you very much for your answer, uncle Olin. I have added more information to my question. –  abdullah kahraman Dec 15 '13 at 13:45

Olin answered your question but I wanted to point something else out. Note that your initial filter Cap, C1 on the 230V side, should be rated to higher than 275V. While voltage rating is typically stated for DC or RMS you really should provide a bit of head room to increase the longevity of the part and for safety when dealing with mains lines. With 230VRMS you should be using something on the order of ~400 or 450V at least to be safe.

Read the answer in AC voltage ratings for capacitors for more information.

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Good point, +1. –  Olin Lathrop Dec 13 '13 at 17:34
This is a 275VAC capacitor. Datasheet is here. –  abdullah kahraman Dec 14 '13 at 11:25
And it must be an X2 type. –  EJP Dec 14 '13 at 23:10

The purpose of the capacitor that connects in parallel to the main is a high frequency filter and should be placed close to the input.

You can read about the capacitor in this application note.

It seems that your question wasn't really about the mains input capacitor C1 but a capacitor connected in the output of the transformer, such a capacitor will not serve any purpose.

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Yes, exactly, my question was about the capacitor that is connected parallel to the secondary (output) of the step-down transformer. Thank you for your answer. –  abdullah kahraman Dec 13 '13 at 13:43

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